Computing a determinant by direct expansion is feasible for 2×2 and 3×3 matrices, but painful for larger ones. The seven properties listed below let you simplify the determinant first , often reducing it to an obvious value with little arithmetic. These properties are also the basis of nearly every JEE-style determinant question.
The seven properties
P1.detAT=detA. The determinant is unchanged by transposition.
Consequence. Every property stated for rows holds also for columns.
P2. If two rows (or columns) of A are interchanged, detA changes sign.
Consequence. If two rows are equal, detA=0 (swapping them gives both the original and its negative).
P3. Multiplying a row (or column) by a scalar λ multiplies the determinant by λ.
Consequence.det(λA)=λndetA for an n×n matrix.
P4.detA=0 if any row (or column) is entirely zero.
P5. Adding a multiple of one row to another does not change the determinant. Same for columns.
Reason. The contribution of the added multiple cancels because of P2 , adding a row to itself would create two equal rows.
P6. Sum-of-rows: if row i of A is a sum u+v, then detA=detAu+detAv, where Au,Av have row i equal to u and v respectively, with all other rows the same.
P7.det(AB)=detA⋅detB. Multiplicative property for matrix products.
Two more useful consequences
If any row is a scalar multiple of another row, detA=0.
det(A−1)=1/detA for invertible A.
Strategic use of properties
Pattern: extracting a common factor. If a row has a common factor, pull it out using P3.
Pattern: simplifying via row addition. Replace a row by itself plus a multiple of another (P5). The determinant is unchanged but the matrix may simplify.
Pattern: detecting linear dependence. If R3=aR1+bR2, then det=0. This is often hidden , look carefully.
A long worked example
Compute detabca+bb+cc+da+2bb+2cc+2d.
C2→C2−C1 (P5): abcbcda+2bb+2cc+2d.
C3→C3−C1 (P5): abcbcd2b2c2d.
C3=2C2 (every entry double). So columns are linearly dependent: det=0.
Vandermonde determinant
det1aa21bb21cc2=(b−a)(c−a)(c−b).
Sketch of proof.C2→C2−C1, C3→C3−C1, then expand along the first row. Each column factors as (b−a) or (c−a). The remaining 2×2 determinant gives (c−b).
This identity generalises: an n×n Vandermonde matrix has determinant ∏i<j(xj−xi).
Worked examples
Example 1. Without expanding, show det125246367=0.
Row 2=2R1. So determinant is zero.
Example 2. Show det1xx21yy21zz2=(y−x)(z−x)(z−y).
Vandermonde formula.
Example 3. Find the determinant of 246369123.
Row 2=2R1, row 3=3R1. So determinant is zero.
Example 4. Show det210432654=0.
R1=2R2+0⋅R3 no...let me check: 2R2=(2,6,10), not R1. Try R1−2R2=(0,−2,−4). Then R1−2R2=−R3. So R1,R2,R3 are linearly dependent, and the determinant is zero.
Example 5. Prove deta−b−c2b2c2ab−c−a2c2a2bc−a−b=(a+b+c)3.
R1→R1+R2+R3: top row becomes (a+b+c,a+b+c,a+b+c). Factor out a+b+c: